# How Do You Simplify the Triple Integral of |xyz| Over an Ellipsoid?

• MattL
In summary, the triple integral can be parametrized using the normal equation for a sphere, with the integral being strictly positive due to the symmetry of the ellipsoid.
MattL
I'm having trouble with evaluating

[Triple Integral] |xyz| dx dy dz

over the region (x/a)^2 + (y/b)^2 + (z/c)^2 <= 1

Do I need to use some sort of parametrisation for the region, and is there some way of dealing with the absolute value function without integrating over the eight octants?

Whilst I've separated the integral into the product of three integrals, I'm not sure if this actually helps?

Well,the function is even and the domain of integration is symmetric wrt the origin,so that would give u a hint upon the limits of integration.The symmetry of the ellipsoid is really useful.

As for the parametrization,i'm sure u'll find the normal one

$$x=a\cos\varphi\sin\vartheta$$

$$y=b\sin\varphi\sin\vartheta$$

$$z=c\cos\vartheta$$

pretty useful.

Daniel.

This integral looks like it's zero with all the symmetry: you know, four positives and four negatives for the integrand. Not sure though as I can't evaluate it. Would like to know though.

saltydog said:
This integral looks like it's zero with all the symmetry: you know, four positives and four negatives for the integrand. Not sure though as I can't evaluate it. Would like to know though.
I think you missed the absolute value sign on the integrand..

arildno said:
I think you missed the absolute value sign on the integrand..

Well . . . no, that's the reason I used for the symmetry but again, I qualify my statements by the fact I can't prove it. For example in the first octant:

$$|xyz|=xyz$$

That's a positive one.

However, in the octant with x<0, y>0 and z>0 we have:

$$|xyz|=-xyz$$

And so forth in the 8 octants leaving 4 positive and 4 negative ones integrated symmetrically (I think).

The integrand is positive almost everywhere; hence, the integral is strictly positive:
Let:
$$x=ar\sin\phi\cos\theta,y=br\sin\phi\sin\theta,z=cr\cos\phi$$
$$0\leq{r}\leq{1},0\leq\theta\leq{2}\pi,0\leq\phi\leq\pi$$
Thus, we may find:
$$dV=dxdydz=abcr^{2}\sin\phi{dr}d\phi{d}\theta$$
$$|xyz|=\frac{abcr^{3}}{2}\sin^{2}\phi|\cos\phi\sin(2\theta)|$$
Doing the r-integrations yield the double integral:
$$I=\frac{(abc)^{2}}{12}\int_{0}^{2\pi}\int_{0}^{\pi}\sin^{3}\phi|\cos\phi\sin(2\theta)|d\phi{d}\theta$$
We have symmetry about $$\phi=\frac{\pi}{2}$$; thus we gain:
$$I=\frac{(abc)^{2}}{24}\int_{0}^{2\pi}|\sin(2\theta)|d\theta$$
We have four equal parts here, and using the part $$0\leq\theta\leq\frac{\pi}{2}$$ yields:
$$I=\frac{(abc)^{2}}{12}$$

Thanks Arildno. MattL, hope I didn't get in your way. I'll go through it to make sure I understand it.

saltydog said:
Thanks Arildno. MattL, hope I didn't get in your way. I'll go through it to make sure I understand it.

No problem.

Think I should be able to give this question a fair go now.

Change variables to x = a*x1, y = b*x2, z = c*x3. The integration region is the unit ball x1^2 + x2^2 + x3^2 <= 1, the integrand is abc*|x1 x2 x3|, and dV = abc * dx1 dx2 dx3. Because of the absolute value and symmetry, the whole integral, I, equals 8 times the integral over the {x1,x2,x3 >= 0} portion of the ball. This gives I = 8abc*int_{x3=0..1} f(x3) dx3, where f(x3) = x3*int_{x1^2 + x2^2 <= 1-x3^2} x1 x2 dx1 dx2. Using polar coordinates (or first integrating over x2 for fixed x1, then integrating over x1) we can easily evaluate f(x3), then integrate it over x3 = 0-->1. The final result is I = abc/6.

R.G. Vickson

## 1. What is a triple integral?

A triple integral is a mathematical tool used to calculate the volume of a three-dimensional object. It involves integrating a function over a three-dimensional region.

## 2. How is a triple integral evaluated?

A triple integral is evaluated by breaking down the three-dimensional region into smaller parts and integrating each part separately. This can be done using either rectangular, cylindrical, or spherical coordinates.

## 3. What are the applications of triple integrals?

Triple integrals have various applications in physics, engineering, and other fields where three-dimensional calculations are needed. For example, they can be used to calculate the mass, center of mass, and moment of inertia of a three-dimensional object.

## 4. What is the difference between a definite and indefinite triple integral?

A definite triple integral has specific limits of integration, while an indefinite triple integral does not. In other words, a definite triple integral will give a numerical value, while an indefinite triple integral will give an equation.

## 5. What are some common mistakes when evaluating triple integrals?

Some common mistakes when evaluating triple integrals include forgetting to change the limits of integration when switching between coordinate systems, miscalculating the bounds of integration, and forgetting to include the appropriate differentials in the integral.

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